How many distinct equilateral triangles can be formed in a regular nonagon having as two of their vertices as the vertices of nonagon ?
Each choice of 2 distinct vertices yields two triangles (one inside the nonagon, and one outside). Since $\displaystyle \dfrac{360}{9}=40$, and $\displaystyle 40\cdot 3=60\cdot 2$, equilateral triangles inside the nonagon among every third vertex use three vertices of the nonagon. So, there are $\displaystyle \binom{9}{2}$ ways to choose two vertices. Among them, there are three distinct sets of three vertices, each three apart. There are $\displaystyle \binom{3}{2}$ ways of choosing each such set of three vertices. Hence, there are $\displaystyle 2\left(\binom{9}{2}-3\binom{3}{2}\right)+3\binom{3}{2}+3 = 66$ distinct equilateral triangles formed where two vertices are vertices of the regular nonagon.
I am unconvinced that this is true, if each equilateral triangle must lie entirely within the nonagon. For example, if we label the vertices counter-clockwise, starting at (1,0), and with the nonagon centered at (0,0), I believe the equilateral triangles whose two vertices are 1 and 5 both lie partially outside the nonagon.
The OP didn't state anything about the triangles lying entirely within the nonagon. The OP required only that the triangles be distinct. Based on that, I assumed the triangles could lie completely inside the nonagon, completely outside the nonagon, or some combination of inside and outside. My wording of inside/outside was only to make a distinction between the two triangles formed by two vertices, and not meant to describe exactly how the triangles lie in relation to the nonagon.
I agree with SlipEternal's analysis for the total number of equilateral triangles. However, if you interpret the "in" in the OP as inside, there are only 21 such:How many distinct equilateral triangles can be formed in a regular nonagon having as two of their vertices as the vertices of nonagon?
Imagine the vertices as being labeled $P_i$ for $0\leq i<9$. Then easily such a triangle with vertices $P_i$ and $P_j$ is inside the nonagon iff $i-j\leq 3$; here $i-j$ is computed mod 9. There are 9 with $i-j=1$, 9 with $i-j=2$ and 3 with $i-j=3$.
I wasn't arguing with your combinatorial analysis. Perhaps the word "in" means something different to you, than it does to me.
I imagined the circumscribed circle of the nonagon, and forming an equilateral triangle (third vertex "towards" the center) generated by the chord formed by vertex 1 and any other point of the circle. It is easy to see that this triangle goes outside the circle (and thus a fortiori outside the nonagon) when we reach vertex 4 (going counter-clockwise), "flips" between vertices 5 and 6, and returns back within the circle at vertex 7.
It is easy to see that we get an equilateral triangle within the nonagon, when the second point is vertex 2, vertex 3, and vertex 4. By rotation, we thus get 27 such triangles, however the triangles (V1, V4,V7), (V2,V5,V8) and (V3,V6,V9) are each counted 3 times (since a rotation of 120 degrees maps these triangles to themselves), so we only obtain 21 distinct triangles.
Let's number the vertices of the nonagon counterclockwise with labels 1 through 9 (by symmetry, it doesn't matter where you start the numbering, it will yield the same result). Vertices 1, 4, and 7 are the vertices of an equilateral triangle. Since we are counting triangles created from two vertices, that triangle is generated by three separate pairs of two vertices: (1 and 4), (4 and 7), and (7 and 1). Rotating the nonagon by 120 degrees produces the mapping of vertices 4 to 1, 7 to 4, and 1 to 7. Hence, the equilateral triangle incident to vertices 1 and 4 is the same triangle as the triangle incident to the mapped vertices 1 and 4. This means that those three triangles are not distinct. Your original post asked for the number of distinct equilateral triangles. Deveno's logic was the rotational mapping gives a way of describing the nonagon's symmetry with respect to equilateral triangles incident to two of the nonagon's vertices.
Hi SlipEternal,
Thanks! But that was not my question. I do understand that the vertices maps on every 120 degrees rotation, so I wasn't questioning that. What I was asking is how Deveno derive 120 degrees in the first place ? Why not 100 degrees or 90 degrees or 130 degrees ? Why specific 120 degrees ? Because each interior angle of a Nonagon is 140 degrees, so how do we relate 140 degrees of nonagon with the 120 degrees rotation ? Or is there some other relation for the mapping ?
Hi SlipEternal,
I think Spock is making a conceptual error. Since the Nonagon is rotating on its axis so the 120 degrees should be considered around the axis of the nonagon which is also the centre of the Nonagon. Now this centre of the nonagon is also the incentre of the equilateral triangle where all the angle bisectors meet. So the area around the centre can be divided into three equal zones since we are dealing with an equilateral triangle and the angle bisectors from the vertices of the triangle meet at the incentre. Hence, each zone is equal to $\displaystyle \frac{360}{3}$ = 120 degrees. Please inform whether my reasoning is correct .
Now each side of the nonagon subtend a 40 degree angle at the centre. So 40 X 3 = 120 degree i.e the triangle remaps at every 3 vertice. Please inform whether my reasoning is correct .
Now I have a question. Is this 120 degrees fixed for any-sided circumscribing polygon, when all the three vertices of the equilateral lie on the vertices of the polygon ? Request advice on this.
Also I have made some observations. All the three vertices of the equilateral triangle lie on the vertices of the polygon only when the external polygon has number of sides which is a multiple of 3. Please inform whether my reasoning is correct.
Request advice on the above. Please inform whether this 120 degrees is fixed for any equilateral triangle within a polygon when all the three vertices lie on the polygon. Please inform soon.